MATH S-21A
Unit 24: Divergence Theorem
Lecture
24.1. There are three integral theorems in three dimensions. We have already seen the fundamental theorem of line integrals and Stokes theorem. The divergence theorem completes the list of integral theorems in three dimensions:
Theorem: Divergence Theorem. If E be a solid bounded by a surface S. The surface S is oriented so that the normal vector points outside. If
~
F be a vector field, then
Z Z Z E div( ~F ) dV = Z Z S ~ F · dS .
24.2. To see why this is true, take a small box [x, x + dx] × [y, y + dy] × [z, z + dz]. The flux of ~F = [P, Q, R] through the faces perpendicular to the x-axes is [ ~F (x + dx, y, z) · [1, 0, 0] + ~F (x, y, z) · [−1, 0, 0]]dydz = P (x + dx, y, z) − P (x, y, z) ∼ Px dxdydz.
Similarly, the flux through the y-boundaries is Py dydxdz and the flux through the
two z-boundaries is Pz dzdxdy. The total flux through the faces of the cube is
(Px + Py + Pz) dxdydz = div( ~F ) dxdydz. A general solid can be approximated as
a union of small cubes. The sum of the fluxes through all the cubes consists now of the flux through all faces without neighboring faces. Important is that fluxes through adjacent sides cancel. The sum of all the fluxes of the cubes is the flux through the boundary of the union. The sum of all the div( ~F ) dxdydz is a Riemann sum approx-imation for the integral RRR
Gdiv( ~F ) dxdydz. In the limit, when dx, dy, dz all go to
Multivariable Calculus
24.3. The theorem explains what divergence means. If we integrate the divergence over a small cube, it is equal the flux of the field through the boundary of the cube. If this is positive, then more field exits the cube than entering the cube. There is field “generated” inside. The divergence measures the “expansion” of the field.
Examples
24.4. Let ~F (x, y, z) = [x, y, z] and let S be the unit sphere. The divergence of ~F is the constant function div( ~F ) = 3 and RRRGdiv( ~F ) dV = 3 · 4π/3 = 4π. The flux through the boundary isRRS~r · ~ru× ~rv dudv =
RR S|~r(u, v)| 2sin(v) dudv = Rπ 0 R2π 0 sin(v) dudv =
4π also. We see that the divergence theorem allows us to compute the area of the sphere from the volume of the enclosed ball or compute the volume from the surface area.
24.5. What is the flux of the vector field ~F (x, y, z) = [2x, 3z2+ y, sin(x)] through the
solid G = [0, 3]×[0, 3]×[0, 3]\([0, 3]×[1, 2]×[1, 2]∪[1, 2]×[0, 3]×[1, 2]∪[0, 3]×[0, 3]×[1, 2]) which is a cube where three perpendicular cubic holes have been removed? Solution: Use the divergence theorem: div( ~F ) = 2 and so RRRGdiv( ~F ) dV = 2RRRG dV = 2Vol(G) = 2(27−7) = 40. Note that the flux integral here would be over a complicated surface over dozens of rectangular planar regions.
24.6. Find the flux of curl(F ) through a torus if ~F = [yz2, z + sin(x) + y, cos(x)] and the torus has the parametrization
~
r(θ, φ) = [(2 + cos(φ)) cos(θ), (2 + cos(φ)) sin(θ), sin(φ)] .
along the boundary, the volume of a region can be computed as a flux integral: Take for example the vector field ~F (x, y, z) = [x, 0, 0] which has divergence 1. The flux of this vector field through the boundary of a solid region is equal to the volume of the solid: RRδG[x, 0, 0] · ~dS = Vol(G).
24.8. How heavy are we, at distance r from the center of the earth?
Solution: The law of gravity can be formulated as div( ~F ) = 4πρ, where ρ is the mass density. We assume that the earth is a ball of radius R. By rotational symmetry, the gravitational force is normal to the surface: ~F (~x) = ~F (r)~x/||~x||. The flux of ~F through a ball of radius r isRRS
r
~
F (x) · ~dS = 4πr2F (r). By the divergence theorem,~ this is 4πMr = 4π
RRR
Brρ(x) dV , where Mr is the mass of the material inside Sr.
We have (4π)2ρr3/3 = 4πr2F (r) for r < R and (4π)~ 2ρR3/3 = 4πr2F (r) for r ≥ R.~
Inside the earth, the gravitational force ~F (r) = 4πρr/3. Outside the earth, it satisfies ~ F (r) = M/r2 with M = 4πR3ρ/3. 0.5 1.0 1.5 2.0 0.2 0.4 0.6 0.8 1.0
Multivariable Calculus
The fundamental theorem for line integrals, Green’s theorem, Stokes theorem and di-vergence theorem are all part of one single theorem RAdF = RδAF , where dF is a exterior derivative of F and where δA is the boundary of A. It generalizes the fundamental theorem of calculus.
Fundamental theorem of line integrals: If C is a curve with boundary {A, B} and f is a function, then
Z
C
∇f · ~dr = f (B) − f (A)
Remarks.
1) For closed curves, the line integral RC∇f · ~dr is zero.
2) Gradient fields are path independent: if ~F = ∇f , then the line integral between two points P and Q does not depend on the path connecting the two points.
3) The line integral theorem holds in any dimension. In one dimension, it reduces to the fundamental theorem of calculus Rabf0(x) dx = f (b) − f (a)
4) The theorem justifies the name conservative for gradient vector fields. 5) The term “potential” was coined by George Green who lived from 1783-1841. 24.10. Example. Let f (x, y, z) = x2 + y4 + z. Find the line integral of the vector field ~F (x, y, z) = ∇f (x, y, z) along the path ~r(t) = [cos(5t), sin(2t), t2] from t = 0 to t = 2π.
Solution. ~r(0) = [1, 0, 0] and ~r(2π) = [1, 0, 4π2] and f (~r(0)) = 1 and f (~r(2π)) = 1 + 4π2. The fundamental theorem of line integral gives R
C∇f ~dr = f (r(2π)) − f (r(0)) =
4π2.
Green’s theorem. If R is a region with boundary C and ~F is a vector field, then Z Z R curl( ~F ) dxdy = Z C ~ F · ~dr . 24.11. Remarks.
1) Greens theorem allows to switch from double integrals to one-dimensional integrals. 2) The curve is oriented in such a way that the region is to the left.
3) The boundary of the curve can consist of piecewise smooth pieces.
4) If C : t 7→ ~r(t) = [x(t), y(t)], the line integral is Rab[P (x(t), y(t)), Q(x(t), y(t))] · [x0(t), y0(t)] dt.
5) Green’s theorem was found by George Green (1793-1841) in 1827 and by Mikhail Ostrogradski (1801-1862).
24.12. Example. Find the line integral of the vector field ~F (x, y) = [x4+ sin(x) +
y, x + y3] along the path ~r(t) = [cos(t), 5 sin(t) + log(1 + sin(t))], where t runs from
t = 0 to t = π.
Solution. curl( ~F ) = 0 implies that the line integral depends only on the end points (0, 1), (0, −1) of the path. Take the simpler path ~r(t) = [−t, 0], −1 ≤ t ≤ 1, which has velocity ~r0(t) = [−1, 0]. The line integral isR1
−1[t
4− sin(t), −t] · [−1, 0] dt = −t5/5|1 −1 =
−2/5.
Remark We could also find a potential f (x, y) = x5/5 − cos(x) + xy + y5/4. It has the property that grad(f ) = F . Again, we get f (0, −1) − f (0, 1) = −1/5 − 1/5 = −2/5.
Stokes theorem. If S is a surface with boundary C and ~F is a vector field, then Z Z S curl( ~F ) · dS = Z C ~ F · ~dr . 24.13. Remarks.
1) Stokes theorem allows to derive Greens theorem: if ~F is z-independent and the surface S is contained in the xy-plane, one obtains the result of Green.
2) The orientation of C is such that if you walk along C and have your head in the direction of the normal vector ~ru × ~rv, then the surface to your left.
3) Stokes theorem was found by Andr´e Amp`ere (1775-1836) in 1825 and rediscovered by George Stokes (1819-1903).
4) The flux of the curl of a vector field does not depend on the surface S, only on the boundary of S.
5) The flux of the curl through a closed surface like the sphere is zero: the boundary of such a surface is empty.
24.14. Example. Compute the line integral of ~F (x, y, z) = [x3+ xy, y, z] along the polygonal path C connecting the points (0, 0, 0), (2, 0, 0), (2, 1, 0), (0, 1, 0) in that order. Solution. The path C bounds a surface S : ~r(u, v) = [u, v, 0] parameterized by R = [0, 2] × [0, 1]. By Stokes theorem, the line integral is equal to the flux of curl( ~F )(x, y, z) = [0, 0, −x] through S. The normal vector of S is ~ru × ~rv = [1, 0, 0] ×
Multivariable Calculus
Divergence theorem: If S is the boundary of a region E in space and ~F is a vector field, then
Z Z Z B div( ~F ) dV = Z Z S ~ F · ~dS . 24.15. Remarks.
1) The divergence theorem is also called Gauss theorem.
2) It is useful to determine the flux of vector fields through surfaces. 3) It can be used to compute volume.
3) It was discovered in 1764 by Joseph Louis Lagrange (1736-1813), later it was redis-covered by Carl Friedrich Gauss (1777-1855) and by George Green.
4) For divergence free vector fields ~F , the flux through a closed surface is zero. Such fields ~F are also called incompressible or source free.
24.16. Example. Compute the flux of the vector field ~F (x, y, z) = [−x, y, z2] through
the boundary S of the rectangular box [0, 3] × [−1, 2] × [1, 2].
Solution. By Gauss theorem, the flux is equal to the triple integral of div(F ) = 2z over the box: R3
0
R2
−1
R2
1 2z dxdydz = (3 − 0)(2 − (−1))(4 − 1) = 27.
24.17. How do these theorems fit together? In n-dimensions, there are n theorems. We have here seen the situation in dimension n = 2 and n = 3, but one could continue. The fundamental theorem of line integrals generalizes directly to higher dimensions. Also the divergence theorem generalizes directly since an n-dimensional integral in n dimensions. The generalization of curl and flux would need more explanation as for n = 4 already, the curl of a vector field is a 6-dimensional object. It is a n(n − 1)/2 dimensional object in general.
1
1
FTC−→1
1
FTL−→2
Green−→1
1
FTL−→3
Stokes−→3
Gauss−→1
field with 4 components into an object with 6 components. Then there is a second curl, which maps an object with 6 components back to a vector field, we would have to look at 1 − 4 − 6 − 4 − 1.
24.19. When setting up calculus in dimension n, one talks about differential forms instead of scalar fields or vector fields. Functions are 0 forms or n-forms. Vector fields can be described by 1 or n − 1 forms. The general formalism defines a derivative d called exterior derivative on differential forms. It maps k forms to k +1 forms. There is also an integration of k-forms on k-dimensional objects. The boundary operation δ which maps a k-dimensional object into a k − 1 dimensional object. This boundary operation is dual to differentiation. They both satisfy the same relation dd(F ) = 0 and δδG = 0. Differentiation and integration are linked by the general Stokes theorem:
Z
δGF =
Z
GdF
24.20. One can see this as a single theorem, the fundamental theorem of multi-variable calculus. The theorem is simpler in quantum calculus, where geometric objects and fields are on the same footing. There are various ways how one can gen-eralize this. One way is to write it as < δG, F >=< G, dF > which in linear algebra would be written as [ATv, w] = [v, Aw], where AT is the transpose of a matrix A [v, w]
is the dot product. Since traditional calculus we deal with ”smooth” functions and fields, we have to pay a prize and consider in turn ”singular” objects like points or curves and surfaces. These are idealized objects which have zero diameter, radius or thickness.
24.21. So, it is all about geometries and fields. Geometries are curves, or surfaces or solids. Fields are scalar functions or vector fields. Geometries G can be “dif-ferentiated” by taking the boundary δG. Fields F can be differentiated by applying differential operators dF like grad, curl or div. And then there is integration which pairs up geometries G and fields F . The fundamental theorem RδGF = RGdF tells that taking the boundary on the object is dual to taking the derivative of the field. 24.22. Nature likes simplicity and elegance 1 and therefore found a quantum mathe-matics to be more fundamental. But a geometry in which geometries and fields are indistinguishable manifests only in the very small.
Multivariable Calculus
Homework This homework is due on Tuesday, 8/3/2021.
Problem 24.1: Find the flux of the vector field ~F (x, y, z) = [xy, yz, zx] through the cylinder x2 + y2 ≤ 1, 0 < z ≤ 2 without the bottom disk
z = 0. Hint: use the divergence theorem by closing the surface first and then computing the flux through the bottom.
Problem 24.2: Compute using the divergence theorem the flux of the vector field ~F (x, y, z) = [9y, 2xy, 4yz+234xy] through the unit cube [0, 1]× [0, 1] × [0, 1].
Problem 24.3: Use the divergence theorem to calculate the flux of ~
F (x, y, z) = [x3, y3, z3] through the sphere S : x2+ y2+ z2 = 1, where the
sphere is oriented so that the normal vector points outwards. Problem 24.4: Assume the vector field
~
F (x, y, z) = [15x3+ 42xy2, y3+ eysin(z), eycos(z) + 15z3] is the magnetic field of the sun whose surface is a sphere of radius 3 oriented with the outward orientation. Compute the magnetic fluxRRSF ·~
~ dS.
Problem 24.5: Find R RSF · ~~ dS, where ~F (x, y, z) = [−150x, 25y, 25z] and S is the boundary of the solid built with the 18 cubes shown in the picture. Each cube has length 1.
